Question

If $$\alpha ,\beta \ne 0,\,{\text{and }}\,f\left( n \right) = {\alpha ^n} + {\beta ^n}$$      and \[\left| \begin{array}{l} \,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\\ 1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\\ 1 + f\left( 2 \right)\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 4 \right) \end{array} \right| = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},\]         then $$K$$ is equal to:

A. 1  
B. $$- 1$$
C. $$\alpha \beta $$
D. $$\frac{1}{{\alpha \beta }}$$
Answer :   1
Solution :
Consider
\[\left| \begin{array}{l} \,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\\ 1 + f\left( 1 \right)\,\,\,\,\,\,\,\,\,1 + f\left( 2 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\\ 1 + f\left( 2 \right)\,\,\,\,\,\,\,\,1 + f\left( 3 \right)\,\,\,\,\,\,\,\,\,\,1 + f\left( 4 \right) \end{array} \right|\]
\[ = \left| \begin{array}{l} \,\,\,1 + 1 + 1\,\,\,\,\,\,\,\,\,\,\,\,\,1 + \alpha + \beta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^2} + {\beta ^2}\\ \,\,1 + \alpha + \beta \,\,\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^2} + {\beta ^2}\,\,\,\,\,\,\,\,\,1 + {\alpha ^3} + {\beta ^3}\\ 1 + {\alpha ^2} + {\beta ^2}\,\,\,\,\,\,1 + {\alpha ^3} + {\beta ^3}\,\,\,\,\,\,\,\,\,\,1 + {\alpha ^4} + {\beta ^4} \end{array} \right|\]
\[ = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right| \times \,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right|\]
\[ = {\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\beta \\ 1\,\,\,\,\,\,\,{\alpha ^2}\,\,\,\,\,\,\,{\beta ^2} \end{array} \right|^2}\]
$$\eqalign{ & = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2} \cr & {\text{So, }}\boxed{k = 1} \cr} $$

Releted MCQ Question on
Algebra >> Matrices and Determinants

Releted Question 1

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$  be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$  be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

A. $$C$$ is empty
B. $$B$$  has as many elements as $$C$$
C. $$A = B \cup C$$
D. $$B$$  has twice as many elements as elements as $$C$$
Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
Releted Question 3

Let $$a, b, c$$  be the real numbers. Then following system of equations in $$x, y$$  and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$     has

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$

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Matrices and Determinants


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